LeetCode 326 Power of Three

Problem:

Given an integer, write a function to determine if it is a power of three.

Could you do it without using any loop / recursion?

Summary:

用非循环/递归的方法判断数n是否为3的整数幂。

Analysis:

1. 循环：将n逐次除以3，判断是否为3的整数幂。

 1 class Solution {
 2 public:
 3     bool isPowerOfThree(int n) {
 4         if (n <= 0) {
 5             return false;
 6         }
 7         while (n > 1) {
 8             if (n % 3) {
 9                 return false;
10             }
11             
12             n /= 3;
13         }
14         
15         return true;
16     }
17 };

2. 递归：若n为3的整数幂，n/3必为3的整数幂，以这个思想构建递归。

 1 class Solution {
 2 public:
 3     bool isPowerOfThree(int n) {
 4         if (n <= 0) {
 5             return false;
 6         }
 7         
 8         if (n == 1) {
 9             return true;
10         }
11         
12         return (n % 3 == 0) && isPowerOfThree(n / 3);
13     }
14 };

3. 非循环/递归：n若为3的整数幂，则 n = 3^log₃n = 3^{log₂n / log ₂3}

1 class Solution {
2 public:
3     bool isPowerOfThree(int n) {
4         return (n > 0) && (n == pow(3, round(log(n) / log(3))));
5     }
6 };